User:Egm6321.f12.team6.Hu.M/report7

Given

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$$ \displaystyle y''+a_0^2y=f(t)$$
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 * (2.1)
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 * Where $$ \displaystyle t\geq 0 $$
 * Initial conditions: $$ \displaystyle y(t_0),y'(t_0) $$
 * Equations in the book:
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$$ \displaystyle \frac{d^2}{dt^2}( \widetilde{\psi})_l(t)+(\mu ^2_1+1)( \widetilde{\psi})_l(t)-(\widetilde{f_k})_l(t)=0 $$
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 * (2.2)
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$$ \displaystyle ( \widetilde{\psi})_l(t)=( \widetilde{\psi})_l(t_n)cos(\lambda _l(t-t_n))+\lambda _l^{-1}\frac{d}{dt}( \widetilde{\psi})_l(t_n)sin(\lambda _l(t-t_n))+\lambda _l^{-1}\int _{t_n}^{t}(\widetilde{f_k})_{l}(s)sin(\lambda _l(t-s))ds $$
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 * (2.3)
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===Problem Statement ===
 * 1.Use variation of parameters to show that:
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$$ \displaystyle y(t)=y_0cos(a_0(t-t_0))+\frac{y'(t_0)}{a_0}sin(a_0(t-t_0))+\frac{1}{a_0}\int_{t_0}^{t}f(\tau)sin(a_0(t-\tau))d\tau $$
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 * (2.4)
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 * 2.Compare Eq2.1, Eq2.4 and the given initial condition with Eq2.2 and Eq2.3 with is from Dong 2012, A trigonometric integrator psendo-spectral discretization of N-coupled nonlinear Klein-Gordon equations, Numerical Algorithms, May.

Nomenclature

 * $$ \displaystyle y'':=\frac{d^2y}{dx^2} $$

Problem 1

 * To find the homogenous solution of
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$$ \displaystyle y''+a_0^2y=0 $$
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 * (2.5)
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 * We assume the trial solution as:
 * $$ \displaystyle y=e^{rx} $$
 * So Eq2.3 can be written as:
 * $$ \displaystyle r^2+a_0^2=0 $$
 * Then we have:
 * $$ \displaystyle r=\pm ia_0 $$
 * Then the homogenous solution of Eq2.1 should be:
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$$ \displaystyle y_H(t)=C_1cos(a_0t)+C_2sin(a_0t) $$
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 * (2.6)
 * }
 * With the given initial condition $$ \displaystyle y(t_0),y'(t_0) $$ we know that:
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$$ \displaystyle y(t_0)=C_1cos(a_0t_0)+C_2sin(a_0t_0) $$
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 * (2.6)
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$$ \displaystyle \frac{y'(t_0)}{a_0}=-C_1sin(a_0t_0)+C_2cos(a_0t_0) $$
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 * (2.8)
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 * From Eq2.7 and Eq2.8 we can easily get:
 * $$ \displaystyle C_1=y(t_0)cos(a_0t_0)-\frac{y'(t_0)}{a_0}sin(a_0t_0) $$
 * $$ \displaystyle C_2=y(t_0)sin(a_0t_0)+\frac{y'(t_0)}{a_0}cos(a_0t_0) $$
 * Then we have the first and the second homogenous solution:
 * $$ \displaystyle u_1(t)=C_1cos(a_0t)=[y(t_0)cos(a_0t_0)-\frac{y'(t_0)}{a_0}sin(a_0t_0)]cos(a_0t) $$
 * $$ \displaystyle u_2(t)=C_1cos(a_0t)=[y(t_0)cos(a_0t_0)-\frac{y'(t_0)}{a_0}sin(a_0t_0)]cos(a_0t) $$
 * $$ \displaystyle u_1(t)+u_2(t)=y(t_0)cos(a_0(t-t_0))+\frac{y'(t_0)}{a_0}sin(a_0(t-t_0)) $$
 * Using variation of parameters with Eq2.1 we can have:
 * $$ \displaystyle h(t)=u_1^2(t) $$
 * Also:
 * $$ \displaystyle y_P(t)=u_1(t)\int \frac{1}{h(t)}[\int h(t)\frac{f(t)}{u_1(t)}dt]dt $$
 * Which can be written as:
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$$ \displaystyle y_P(t)=\frac{1}{a_0}\int _{t_0}^{t}f(\tau)sin(a_0(t-\tau )) d\tau $$
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 * (2.9)
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 * With Eq2.8 and Eq2.9 we have the complete solution for Eq2.1:
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$$ \displaystyle y(t)=y(t_0)cos(a_0(t-t_0))+\frac{y'(t_0)}{a_0}sin(a_0(t-t_0))+\frac{1}{a_0}\int _{t_0}^{t}f(\tau )sin(a_0(t-\tau ))d\tau $$
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 * (2.10)
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Problem 2

 * Compare Eq2.1 and Eq2.4 with Eq2.2 and Eq2.3, it is obvious that:
 * $$ \displaystyle y(t)=(\widetilde{\psi})_l(t) $$
 * $$ \displaystyle f(t)=(\widetilde{f_k})_l(t) $$
 * $$ \displaystyle a_0^2=\lambda _l^2=\mu _1^2+1 $$